SELECT DISTINCT user_id support_date WHERE 中的最高值

Question

tb_supporters:

user_id          support_date
-----------------------------
   1              1301283373
   2              1301283743
   3              1301799207
   1              1403862904
   2              1405174895
   3              1415266390

作为查询结果，它应该输出每个唯一user_id中最高的support_date。

我的实际查询：

"SELECT DISTINCT(user_id), support_date
 FROM tb_supporters WHERE
 support_date < " . (time() - 60*60*24*7)

但是，它并不是 select除 user_id 之外最高的 support_date。尝试了 MAX(support_date)、GROUP BY user_id、ORDER BY support_date DESC，但没有任何帮助，它仍然是 select 最低的 support_date。

请你引导我进入正确的方向如何完成这个任务 select 只有每个 user_id 旁边最高的 support_date？

结果应该是：

user_id          support_date
-----------------------------
   1              1403862904
   2              1405174895
   3              1415266390

我查询中的 WHERE 子句是为了排除 user_ids，其中 support_date 在过去 7 天的范围内。

重要说明： 在这 7 天的范围内存在 user_id 我根本不想被 select 编辑，即使是他们最高 support_date.

Answer 1

样本Table

CREATE Table table_sample(id INT, date INT);

一些 INSERT 的

INSERT INTO table_sample VALUES(1,1301283373);
INSERT INTO table_sample VALUES(2,1301283743);
INSERT INTO table_sample VALUES(3,1301799207);
INSERT INTO table_sample VALUES(1,1403862904);
INSERT INTO table_sample VALUES(2,1405174895);
INSERT INTO table_sample VALUES(3,1415266390);

SELECT * FROM table_sample;
+------+------------+
| id   | date       |
+------+------------+
|    1 | 1301283373 |
|    2 | 1301283743 |
|    3 | 1301799207 |
|    1 | 1403862904 |
|    2 | 1405174895 |
|    3 | 1415266390 |
+------+------------+
6 rows in set (0.00 sec)

Your query 

SELECT id,MAX(date) FROM table_sample GROUP BY id;
+------------+------+
|  id |MAX(date)  |
+------------+------+
| 1   |  1403862904 |
| 2   |  1405174895 |
| 3   |  1415266390 |
+------------+------+
3 rows in set (0.00 sec)

要查找日期在过去 7 天的范围内，您可以

SELECT id,MAX(date) FROM table_sample WHERE date <= (UNIX_TIMESTAMP() - (60 * 60 * 24 * 7)) GROUP BY id;
+------+------------+
| id   | MAX(date)  |
+------+------------+
|    1 | 1403862904 |
|    2 | 1405174895 |
|    3 | 1415266390 |
+------+------------+
3 rows in set (0.00 sec)

Answer 2

试试这样的东西：

SELECT user_id,max(support_date) as support_date 
FROM mytable 
GROUP BY user_id    
ORDER BY support_date DESC

Answer 3

SELECT id, max(date) FROM tb_supporters group by id;

Answer 4

您也可以使用默认的 MySQL unix 时间戳函数

SELECT 
    user_id, MAX(support_date)
FROM
    tb_supporters
WHERE
    support_date < (UNIX_TIMESTAMP() - (60 * 60 * 24 * 7))
GROUP BY user_id

Answer 5

The WHERE clause in my query is to keep out user_ids where the support_date is inside a range of the past 7 days.

你可以这样改写：

The WHERE clause in my query is to select user_ids where highest support_date is outside a range of the past 7 days.

您可以将 GROUP BY 和 MAX 与 HAVING 子句一起使用：

SELECT user_id, MAX(support_date)
FROM tb_supporters
GROUP BY user_id
HAVING MAX(support_date) < UNIX_TIMESTAMP(CURRENT_TIMESTAMP - INTERVAL 7 DAY)

请注意，您不能为此使用 WHERE 子句，因为它会删除行用户支持在过去 7 天内；不是组.

SELECT DISTINCT user_id support_date WHERE 中的最高值

SELECT DISTINCT user_id of highest value in support_date WHERE

mysql

max

distinct

sql-order-by

where